Question

The sum of first $n$ terms of the series
$1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+....$ is $\frac{n(n+1{)}^2}{2}$ when $n$ is even. When $n$ is odd the sum is

• A. $\frac{n^2}{2}(n+1)$
• B. $\frac{n}{2}(n-1{)}^2$
• C. $\frac{n}{2}(n-1)$
• D. ${\frac{n(n+1)}{2}}^2$

Answer 1

The correct option is B $\frac{n}{2}(n-1{)}^2$

When n is odd, let $n=2m+1$

∴ The req. sum

$=1^2+2\times 2^2+3^2+2\times 4^2+\dots ..+2(2m{)}^2+(2m+1)2$

$=\sum (2m+1{)}^2+4[1^2+2^2+3^2+\dots .+m^2]$

$=\frac{(2m+1)(2m+2)(4m+2+1)}{6}+\frac{4m(m+1)(2m+1)}{6}$

$=\frac{(2m+1)(m+1)}{6\left[2\right(4m+3)+4m]}$

$=\frac{(2m+1)(2m+2)(6m+3)}{6}=\frac{\left(\right(2m+1{)}^2(2m+2)}{2}$

$=\frac{n^2(n+1)}{2}[\because 2m+1=n]$